A mixed ether (P), when heated with excess of hot concentrated hydrogen iodide produces two different alkyl iodides which when treated with aq. NaOH give compounds (Q) and (R). Both (Q) and (R) give yellow precipitate with NaOI. Identify the mixed ether (P):

We work backwards. Both (Q) and (R) give a positive iodoform test (yellow precipitate with NaOI). Since (Q) and (R) are obtained by treating alkyl iodides with aq. NaOH, they must be alcohols. Alcohols that give a positive iodoform test must possess the structural unit CH3-CH(OH)-. Therefore, both alkyl iodides must contain the CH3-CH(I)- group. The precursor mixed ether (P) was treated with excess concentrated HI to yield these two alkyl iodides. For both sides of the ether to produce iodides capable of forming secondary alcohols with terminal methyls upon hydrolysis, the ether must be composed of two branches that look like -CH(CH3)-. Let's evaluate the correct option (1): Structure (P) is an ether linking a sec-butyl group and a secondary-pentyl group (or similar structure). Excess HI cleaves the ether to give two secondary alkyl iodides. Hydrolysis with aq. NaOH converts both alkyl iodides into secondary alcohols. Both secondary alcohols have a methyl group directly adjacent to the hydroxyl-bearing carbon, fully satisfying the iodoform test requirement. Excess hot HI cleaves ethers (R-O-R') into two moles of alkyl iodides (R-I and R'-I). Only secondary alcohols with a methyl group or ethanol itself give positive iodoform tests. Look for an ether containing CH3-CH(O-)- branching on BOTH sides. Class 12 Chemistry: Alcohols, Phenols and Ethers Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

A mixed ether (P), when heated with excess of hot concentrated hydrogen iodide produces two different alkyl iodides which when treated with aq. NaOH give compounds (Q) and (R). Both (Q) and (R) give y

We work backwards. Both (Q) and (R) give a positive iodoform test (yellow precipitate with NaOI). Since (Q) and (R) are obtained by treating alkyl iodides with aq. NaOH, they must be alcohols. Alcohols that give a positive iodoform test must possess the structural unit CH3-CH(OH)-. Therefore, both alkyl iodides must contain the CH3-CH(I)- group. The precursor mixed ether (P) was treated with excess concentrated HI to yield these two alkyl iodides. For both sides of the ether to produce iodides c

A mixed ether (P), when heated with excess of hot concentrated hydrogen iodide produces two different alkyl iodides which when treated with aq. NaOH give compounds (Q) and (R). Both (Q) and (R) give yellow precipitate with NaOI. Identify the mixed ether (P):

We work backwards. Both (Q) and (R) give a positive iodoform test (yellow precipitate with NaOI). Since (Q) and (R) are obtained by treating alkyl iodides with aq. NaOH, they must be alcohols. Alcohols that give a positive iodoform test must possess the structural unit CH3-CH(OH)-. Therefore, both alkyl iodides must contain the CH3-CH(I)- group. The precursor mixed ether (P) was treated with excess concentrated HI to yield these two alkyl iodides. For both sides of the ether to produce iodides capable of forming secondary alcohols with terminal methyls upon hydrolysis, the ether must be composed of two branches that look like -CH(CH3)-. Let's evaluate the correct option (1): Structure (P) is an ether linking a sec-butyl group and a secondary-pentyl group (or similar structure). Excess HI cleaves the ether to give two secondary alkyl iodides. Hydrolysis with aq. NaOH converts both alkyl iodides into secondary alcohols. Both secondary alcohols have a methyl group directly adjacent to the hydroxyl-bearing carbon, fully satisfying the iodoform test requirement. Excess hot HI cleaves ethers (R-O-R') into two moles of alkyl iodides (R-I and R'-I). Only secondary alcohols with a methyl group or ethanol itself give positive iodoform tests. Look for an ether containing CH3-CH(O-)- branching on BOTH sides. Class 12 Chemistry: Alcohols, Phenols and Ethers Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

A mixed ether (P), when heated with excess of hot concentrated hydrogen iodide produces two different alkyl iodides which when treated with aq. NaOH give compounds (Q) and (R). Both (Q) and (R) give y

We work backwards. Both (Q) and (R) give a positive iodoform test (yellow precipitate with NaOI). Since (Q) and (R) are obtained by treating alkyl iodides with aq. NaOH, they must be alcohols. Alcohols that give a positive iodoform test must possess the structural unit CH3-CH(OH)-. Therefore, both alkyl iodides must contain the CH3-CH(I)- group. The precursor mixed ether (P) was treated with excess concentrated HI to yield these two alkyl iodides. For both sides of the ether to produce iodides c

A mixed ether (P), when heated with excess of hot concentrated hydrogen iodide produces two different alkyl iodides…

Year	2026	2025	2024	Total
Questions	7	5	11	23

Q66 jee_main_2026_24_january_evening Classification of Alcohols

From the following, how many compounds contain at least one secondary alcohol? (I)

Classification of Alcohols diagram for Q66 - JEE Main 2026 Evening — Various organic structures displaying alcohol functional groups.

(II)

(III)

(IV)

(V)

(VI)

Choose the correct answer from the options given below:

A. Five
B. Three
C. Four
D. two

Solution

Core Logic

A secondary alcohol features a hydroxyl group (-OH) attached to a carbon atom that is bonded to two other carbon atoms ( $2°$ $2^{\circ}$ carbon). Evaluating the structures based on visual analysis: Compounds II, IV, and V contain at least one hydroxyl group on a $2°$ $2^{\circ}$ carbon. Therefore, 3 compounds feature a secondary alcohol.

Pattern Recognition

Spot the carbon carrying the -OH group. Count the C-C bonds attached to it. 1 = Primary, 2 = Secondary, 3 = Tertiary. Phenols are aromatic and distinct from secondary alcohols.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers

Q54 jee_main_2026_28_january_morning Kolbe Schmitt Reaction

Consider the following reaction sequence Compound (x) [76.6%C, 6.38%H, vapour density 47]

(i)CO₂,NaOH,120°C,high pressure (ii)H₃O⁺

Compound (y) (Major Product) Compound (y) develops characteristic colour with neutral

FeCl₃

solution. Identify the INCORRECT statement from the following for the above sequence.

A. Both compounds x and y will dissolve in NaOH.
B. Compound y will dissolve in $NaHCO₃$ $\mathrm{NaHCO}_3$ and evolve a gas.
C. Compound x is more acidic than compound y.
D. Both compounds x and y will burn with sooty flame.

Solution

Core Logic

Based on empirical data for (x): $Molar Mass = 2 \times Vapour Density = 2 \times 47 = 94~g/mol$ . $C = 76.6% (76.6/12) \approx 6.38$ $H = 6.38% (6.38/1) \approx 6.38$ $O = 17.02% (17.02/16) \approx 1.06$ Ratio $C:H:O \approx 6:6:1 C₆H₆O$ (Phenol).

Step 1: Identifying Compounds

Compound (x) is Phenol. It reacts with $CO₂/NaOH$ $\mathrm{CO}_2/\mathrm{NaOH}$ via Kolbe-Schmitt reaction to form Salicylic Acid as the major product (Compound y).

Kolbe Schmitt Reaction of Phenol to Salicylic Acid

Step 2: Checking Options

(1) Both Phenol and Salicylic acid dissolve in $NaOH$ $\mathrm{NaOH}$. (True) (2) Salicylic acid (y) has a $-COOH$ $-COOH$ group and dissolves in $NaHCO₃$ $\mathrm{NaHCO}_3$, evolving $CO₂$ $\mathrm{CO}_2$ gas. (True) (3) Salicylic acid (y) is much more acidic than Phenol (x) due to the presence of the carboxylic acid group and intramolecular hydrogen bonding in its conjugate base. Therefore, the statement "x is more acidic than y" is INCORRECT. (4) Being highly aromatic compounds, both burn with a sooty flame. (True)

Pattern Recognition

The reaction sequence $Phenol + CO₂/NaOH arrow Salicylic acid$ is a staple (Kolbe's Reaction). Always test functional group properties: Phenols dissolve in $NaOH$ $\mathrm{NaOH}$ but not $NaHCO₃$ $\mathrm{NaHCO}_3$, while Carboxylic acids dissolve in both.

Chapter Mix

Class 12 Chemistry: Alcohols Phenols and Ethers Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

Q jee_main_2025_07_april_morning Acidity of Phenols

Which of the following compounds is least likely to give effervescence of

CO₂

in presence of aq.

NaHCO₃

?

A.
B.
C. $Ph - NH₃(+) Cl(-)$ $\mathrm{Ph - NH_3^{(+)} Cl^{(-)}}$
D.

Solution

Core Logic

For a compound to release $CO₂$ $\mathrm{CO}_2$ upon reaction with aqueous $NaHCO₃$ $\mathrm{NaHCO}_3$, its acidity must be strictly greater than that of carbonic acid ( $H₂CO₃$ $\mathrm{H_2CO_3}$).

Let's evaluate the acidity of the options:

Picric Acid (2,4,6-trinitrophenol - Option A): Extremely acidic ( $pKₐ \approx 0.38$ $pK_a \approx 0.38$) due to three strong electron-withdrawing nitro groups. Reacts with $NaHCO₃$ $\mathrm{NaHCO}_3$ easily.
4-Nitrobenzoic Acid (Option B): Carboxylic acids generally have $pKₐ \approx 4--5$ $pK_a \approx 4\text{--}5$. More acidic than carbonic acid ( $pKₐ \approx 6.3$ $pK_a \approx 6.3$). Reacts with $NaHCO₃$ $\mathrm{NaHCO}_3$.
Anilinium Chloride (Option 3): A salt of a strong acid and weak base. The anilinium ion is quite acidic ( $pKₐ \approx 4.6$ $pK_a \approx 4.6$) and easily decomposes $NaHCO₃$ $\mathrm{NaHCO}_3$.
4-Nitrophenol (Option D): Only one nitro group is present. Its acidity ( $pKₐ \approx 7.15$ $pK_a \approx 7.15$) is weaker than carbonic acid. Hence, it does not react with sodium bicarbonate to yield effervescence of $CO₂$ $\mathrm{CO}_2$.

Pattern Recognition

Acid-bicarbonate test shortcut:

All carboxylic acids and picric acid give a positive bicarbonate test.
Normal phenols and mono/di-nitrophenols are too weak to decompose bicarbonate.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers Class 12 Chemistry: Carboxylic Acids

Q31 jee_main_2025_07_april_morning Physical Properties of Ethers

Given below are two statements: Statement I: Dimethyl ether is completely soluble in water. However, diethyl ether is soluble in water to a very small extent. Statement II: Sodium metal can be used to dry diethyl ether and not ethyl alcohol. In the light of given statements, choose the correct answer from the options given below:

A. $Statement I is false but Statement II is true$
B. $Both Statement I and Statement II are false$
C. $Statement I is true but Statement II is false$
D. $Both Statement I and Statement II are true$

Solution

Core Logic

Statement I: Dimethyl ether ( $CH₃OCH₃$ $CH_3OCH_3$) is highly soluble in water because its smaller alkyl chain allows substantial hydrogen bonding with water molecules. In contrast, diethyl ether has a larger hydrophobic ethyl group which drastically reduces its water solubility (to about $7.5 g$ $7.5\text{ g}$ per $100 mL$ $100\text{ mL}$). Thus, Statement I is true.

Statement II: Sodium metal ( $Na$ $Na$) reacts violently with alcohols like ethyl alcohol to release hydrogen gas:

2C₂H₅OH + 2Na arrow 2C₂H₅ONa + H₂

Since diethyl ether has no active acidic hydrogen, it does not react with sodium metal. Hence, sodium can dry diethyl ether but cannot be used for ethyl alcohol. Statement II is true.

Pattern Recognition

Ethers are miscible with water primarily when the non-polar alkyl parts are very small. Active hydrogen presence ( $-OH$ $-\text{OH}$ group in alcohols) prevents the use of alkali metals like sodium for moisture removal.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers

Q jee_main_2025_08_april_evening Williamson Ether Synthesis

Which one of the following reactions will not lead to the desired ether formation in major proportion? (Given labels:

iso-Bu \Rightarrow isobutyl

,

sec-Bu \Rightarrow sec-butyl

,

n-Pr \Rightarrow n-propyl

,

^tBu \Rightarrow tert-butyl

,

Et \Rightarrow ethyl

)

A. $^tBuO^-Na^+ + EtBr ^tBu-O-Et$
B. $PhO^-Na^+ + CH₃Br Ph-O-CH₃$
C. $PhO^-Na^+ + n-PrBr n-Pr-O-Ph$
D. $iso-BuO^-Na^+ + sec-BuBr sec-Bu-O-iso-Bu$

Solution

Core Logic

Williamson Ether Synthesis operates strictly via an $SN2$ $S_N2$ mechanism. To optimize ether formation yields, the alkyl halide component MUST be primary ( $1^°$ $1^\circ$) or methyl to evade spatial shielding restrictions.

Let us review the alkyl halide across the options:

Option 1: Ethyl bromide ( $EtBr$ $\text{EtBr}$) is $1^°$ $1^\circ$ $arrow$ $\rightarrow$ Excellent ether yield via substitution.
Option 2: Methyl bromide ( $CH₃Br$ $\text{CH}_3\text{Br}$) is a unhindered methyl halide $arrow$ $\rightarrow$ High substitution efficiency.
Option 3: n-Propyl bromide ( $n-PrBr$ $\text{n-PrBr}$) is $1^°$ $1^\circ$ $arrow$ $\rightarrow$ Clean substitution pathway.
Option 4: sec-Butyl bromide ( $sec-BuBr$ $\text{sec-BuBr}$) is a secondary ( $2^°$ $2^\circ$) alkyl halide. When a sterically hindered $2^°$ $2^\circ$ halide is treated with a strongly basic alkoxide reagent like isobutoxide, alkene elimination ( $E2$ $E2$) competes aggressively and dominates as the major pathway over nucleophilic substitution ( $SN2$ $S_N2$).
Competing elimination reaction mechanism layout for Q37

Pattern Recognition

Williamson Synthesis Rule: Alkoxide can be as massive and complex as desired ( $3^°$ $3^\circ$ or branched), but the Alkyl Halide MUST be unhindered ( $1^°$ $1^\circ$ or methyl). If the halide is $2^°$ $2^\circ$ or $3^°$ $3^\circ$, elimination wins, forming an alkene instead of an ether.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers

Solution & Explanation

Core Logic

Pattern Recognition

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Pattern Recognition

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